Leetcode: Simplify Path

Question

Given an absolute path for a file (Unix-style), simplify it.

For example,

path = “/home/”, => “/home”

path = “/a/./b/../../c/”, => “/c”

click to show corner cases.

Corner Cases:

Did you consider the case where path = “/../”?

In this case, you should return “/”.

Another corner case is the path might contain multiple slashes ‘/’ together, such as “/home//foo/”.

In this case, you should ignore redundant slashes and return “/home/foo”.

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My first try

[code]class Solution(object):
    def simplifyPath(self, path):
        """
        :type path: str
        :rtype: str
        """

        if path=='':
            return ''

        path = path.split('/')
        vis = [True]*len(path)

        for ind in range(len(path)):
            if path[ind]=='' or path[ind]=='.':
                vis[ind]=False
            if path[ind]=='..':
                vis[ind]=False
                if ind>0:
                    vis[ind-1] = False

        temp = [ path[ind] for ind in range(len(path)) if vis[ind]==True]
        temp = '/'.join(temp)

        return '/'+temp

Error

[code]Input:
"/a/./b/../../c/"
Output:
"/a/c"
Expected:
"/c"

My Solution

time complexity: O(n)

space complexity: O(n)

[code]class Solution(object):
    def simplifyPath(self, path):
        """
        :type path: str
        :rtype: str
        """

        if path=='':
            return ''

        path = path.split('/')
        vis = [True]*len(path)

        for ind in range(len(path)):
            if path[ind]=='' or path[ind]=='.':
                vis[ind]=False
            if path[ind]=='..':
                vis[ind]=False
                for tind in range(ind-1,-1,-1):
                    if vis[tind]==True:
                        vis[tind]=False
                        break

        temp = [ path[ind] for ind in range(len(path)) if vis[ind]==True]
        temp = '/'.join(temp)

        return '/'+temp

Other’s Solution

Get idea from here, here2.